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\begin{document}
\title{Problem set 2}
\author{Due date: Feb 19}
\maketitle

\begin{enumerate}
\item Suppose that $T$ is a symmetric bounded operator. Then show that
$$
\|T \| = \sup\{|(Tf,f)|, \quad \|f \|=1 \} \, .
$$
Hint: You may assume the polarization identity
$$
(Tf,g) = \frac{1}{4}[ (T(f+g),f+g) - (T(f-g),f-g) +i (T(f+ig),f+ig) -i (T(f-ig),f-ig) ]
$$
\item Suppose that $G$ is a compact set in $\R^{n}$. Suppose that 
$$
T[f](x) = \int_{G} K(x,y) f(y) dy \, ,
$$
where $K:G\times G \to \R$ is a continuous function for all $x,y \in G$ except for $x=y$. 
Furthermore, suppose that $K$ satisfies
$$
|K(x,y)| \leq \frac{C}{|x-y|^{\alpha}} \, , 
$$
where $\alpha>0$. Find the range of values of $\alpha$ for which the operator $T: \bL^{2}(G) \to \bL^{2}(G)$ is compact.
Hint: Integral operators with continuous kernels are compact, and the norm limit of compact operators is compact.
\item Consider the operator $T: \bL^{2}([0,1]) \to \bL^{2}([0,1])$ defined by
$$
T[f](t) = t\cdot f(t)
$$
\begin{enumerate}
\item Prove that $T$ is a bounded linear operator with $T=T^{*}$, but that $T$ is not compact
\item However, show that $T$ has no eigenvectors
\end{enumerate}
The multiplication operator defined above is shown to have a critical role in the design of quadratures (see , for example).
\item Let $\cH$ be a Hilbert space with basis $\{ e_{k} \}_{k=1}^{\infty}$. Verify that the operator $T$ defined by
$$
T(e_{k}) = \frac{e_{k+1}}{k} \, ,
$$
is compact, but has no eigenvectors
\item Let $\cH$ be a Hilbert space with basis $\{ e_{k} \}_{k=1}^{\infty}$. Verify that the operator $T$ defined by
$$
T(e_{k}) =  \lambda_{k} e_{k} \, ,
$$
is compact if and only if $\lim_{k \to \infty} \lambda_{k} \to 0$.
\item Let $\sigma(T)$ denote the spectrum of a compact operator $T: \cH \to \cH$. Show that
$\lambda \in \sigma(T)$ if and only if $\overline{\lambda} \in \sigma(T^{*})$
\item Let $K$ be a Hilbert-Schmidt kernel which is real and symmetric, i.e. $K:[0,1]\times[0,1]\to \R$ satisfies
$K(x,y)=K(y,x)$ and $K \in \bL^{2}([0,1]\times[0,1])$. Let $T:\bL^{2}([0,1]) \to \bL^{2}([0,1])$ be defined by
$$
T[f](x) = \int_{0}^{1}K(x,y) f(y) dy \, .
$$
Let $\phi_{k}(x)$ be the eigenvectors (with eigenvalues $\lambda_{k}$) that diagonalize $T$. Then:
\begin{enumerate}
\item $\sum_{k} |\lambda_{k}|^{2} < \infty$
\item $K(x,y) = \sum_{k=1}^{\infty} \lambda_{k} \phi_{k}(x) \phi_{k}(y)$
\item Suppose $\tilde{T}$ is an operator which is compact and symmetric. Then $\tilde{T}$ is of Hilbert-Schmidt type if and only if 
$\sum_{n} |\lambda_{n}|^2 < \infty$, where $\{ \lambda_{n} \}$ are the eigenvalues of $\tilde{T}$ counted according to their
multiplicities
\end{enumerate}
\item Let $\cH$ be a Hilbert space.
\begin{enumerate}
\item If $T_{1},T_{2}: \cH \to \cH$ are compact symmetric operators which commute, i.e. $(T_{1} T_{2} = T_{2} T_{1})$, show
that they can be diagonalized simultaneously. In other words, there exists an orthonormal basis for $\cH$ which consists of 
eigenvectors for both $T_{1}$ and $T_{2}$.
\item A linear operator on $\cH$ is normal if $TT^{*} = T^{*} T$. Prove that if $T$ is normal and compact, then $T$ can
be diagonalized.
\item If $U$ is unitary, and $U = \lambda I - T$, where $T$ is compact, then $U$ can be diagonalized.
\end{enumerate}
\item Fredholm theory for non-zero index operators.
An operator $R$ is called a regularizer of an operator $K$ if $R$ is bounded and 
$RK = I - A_{\ell}$ and $KR = I- A_{r}$, where $A_{\ell},A_{r}$ are compact.
\begin{enumerate}
\item Suppose that $K:\cH \to \cH$, and $R$ is a regularizer of $K$, then $\text{dim}\{ \cN(K) \} < \infty$  and
$\text{dim}\{ \cN(R) \} < \infty$
\item If $RK = I - A$, where $A$ is compact, show that $\phi - A \phi = Rf$ has a solution for every $f \in \cN(K^{\ast})^{\perp}$
\item Now further assume that $N(I-A) = \{ 0 \}$. Suppose that $S = (I-A)^{-1}$. 
Show that $\text{Ran}\left((I-KSR)\right) \subset \cN(R)$ and that
$\text{Ran}\left((I-KSR)^{\ast}\right) \subset \cN(K^{\ast})$. 
Combine the previous result and these results to show that $\phi = SRf$ also satisfies $K\phi = f$ as long as 
$f \in N(\cK^{\ast})^{\perp}$.
\item (optional, no extra credit) Show that $\text{Ran(K)} = \cN(K^{*})^{\perp}$ for any operator $K$ which has a
regularizer
\end{enumerate}
\end{enumerate}
\end{document}